Question

at a schools open house, t - shirts and sweatshirts were sold. each item was purchased by either a student or a parent. the two - way frequency table summarizes a random sample of 50 items sold that night. \

$$\begin{tabular}{|c|c|c|} \\hline & student & parent \\\\ \\hline t - shirt & 8 & 19 \\\\ \\hline sweatshirt & 12 & 11 \\\\ \\hline \\end{tabular}$$

let t - shirt be the event that an item randomly chosen from among those sold was a t - shirt. let student be the event that an item randomly chosen from among those sold was purchased by a student. find the following probabilities. write your answers as decimals. (if necessary, consult a list of formulas.) (a) $p(student) = \square$ (b) $p(t - shirt \text{ and } student) = \square$ (c) $p(t - shirt \mid student) = \square$

Explanation:

Step1: Calculate P(student)

First, find the total number of items sold, which is 50 (given). Then, find the number of items purchased by students: 8 (T - shirt, student) + 12 (Sweatshirt, student) = 20.
The formula for probability is \( P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).
So, \( P(\text{student})=\frac{20}{50} = 0.4 \).

Step2: Calculate P(T - shirt and student)

The number of favorable outcomes (T - shirt and student) is 8. The total number of outcomes is 50.
Using the probability formula, \( P(\text{T - shirt and student})=\frac{8}{50}=0.16 \).

Step3: Calculate P(T - shirt | student)

The formula for conditional probability is \( P(A|B)=\frac{P(A\cap B)}{P(B)} \). We know \( P(A\cap B) = P(\text{T - shirt and student}) = 0.16 \) and \( P(B)=P(\text{student}) = 0.4 \).
So, \( P(\text{T - shirt | student})=\frac{0.16}{0.4}=0.4 \).

Answer:

s:
(a) \( \boldsymbol{0.4} \)
(b) \( \boldsymbol{0.16} \)
(c) \( \boldsymbol{0.4} \)